It is known that the complex numbers z r cos isin r, R 1 calculate z 2 and z 3 respectively and dedu

Known complex numbers. z=r(cosθ+isinθ)

z^2=r^2(cosθ+isinθ)^2r^2(cos^2θ-sin^2θ+isin2θ)r^2(cos2θ+isin2θ)

z^3=z*z^2=r(cosθ+isinθ)*r^2(cos2θ+isin2θ)

r^3(cosθcos2θ+isin2θcosθ+isinθcos2θ-sinθsin2θ)

r^3(cos3θ+isin3θ)

It can be summed up from this.

z^n=r^n(cosnθ+isinnθ)√3+i)^7

2^7(√3/2+1/2i)^7

2^7(cosπ/6+isinπ/6)^72^7(cos7π/6+isin7π/6)2^7(-(3/2-1/2i)

2^6(√3+i)

1)z 2=r 2* (cos ) 2+i(sin ) 2)z 3=r 3* (cos ) 3+i(sin ) 3)2) to summarize z 7=r 7*((cos ) 7+i(sin ) 7).

cos = root number 3

sin =1 and do the math yourself.

z^2=r^2(cos2θ+isin2θ),z^3=r^3(cos3θ+isin3θ),z^n=r^n(cosnθ+isinnθ);

Root number 3) + (i)) 7 = 2 7 [cos(7 30 degrees) + isin(7 30 degrees)] = 128 (cos210 degrees + isin210 degrees).

64 [-root number 3)-(i)].

Let Z=cos2a+isin2a

r=(z+1) (z-1)=(cosa+isina) (sina+icosa)=e (i*(-pi bridge bucket type minchai pin tour2))=i

re(r)=0

Let z=a+bi (a,b r).

Then Gao Hu|(a+1+bi)/(a-1+bi)|=1 square to get 2a + 1 + 2bi = 0

Therefore, Qi Yuban a=-1 2, b=0

At this point z=-1 2

and z+2 z r, satisfying the virtual slip.

Therefore z=-1 2

z 7=1 directly gives z=e (2k i 7) k=0,1,2,..6 In its code, k=0 corresponds to z=1, so k=0 should be rounded

Then substitute z+z 2+z 4=e (2k i 7)+e (4k i 7)+e (8k i 7).

Let z=a+bi (a,b r).

then |(a+1+bi)/(a-1+bi)|=1 square to get 2a + 1 + 2bi = 0

Therefore a=-1 2, b=0

At this point z=-1 2

and z+2 z r, satisfied.

Therefore z=-1 2

Let the old companion z=a+bi (a,b r) then |(a+1+bi)/(a-1+bi)|=1 squared to obtain 2a+1+2bi=0, so a=-1 2, b=0, at this time, z=-1 2 and z+2 z r, satisfied, so it is known that z=-1 is open-minded 2

The distance from z to (1,0) is equal to the distance from z to (3,0) for a handstool.

z is on a straight line x=2.

Let the orange potato bright z=2+bi, b r

Because |z|=√4+b²)=3

So b = 5, b = 5

z=2+5i or z=2- circle width 5i

Solve a system of equations.

For reference, please smile at the beams. Oak World.